1181: For Continuous Map Between Topological Spaces, Image of Closure of Subset Is Contained in Closure of Image of Subset

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description/proof of that for continuous map between topological spaces, image of closure of subset is contained in closure of image of subset

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of continuous map.
• The reader knows a definition of closure of subset of topological space.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the image of the closure of any subset of the domain is contained in the closure of the image of the subset.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$S_1$: $\subseteq T_1$
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Statements:
$f(\overline{S_1})\subseteq \overline{f(S_1)}$
//

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2: Proof

Whole Strategy: Step 1: let $p\in \overline{S_1}$ be any and see that $p\in S_1$ or $p\in \overline{S_1}\setminus S_1$; Step 2: when $p\in S_1$, see that $f(p)\in f(S_1)\subseteq \overline{f(S_1)}$; Step 3: when $p\in \overline{S_1}\setminus S_1$, see that $f(p)\in \overline{f(S_1)}$.

Step 1:

Let $p\in \overline{S_1}$ be any.

$p\in S_1$ or $p\in \overline{S_1}\setminus S_1$.

Step 2:

Let us suppose that $p\in S_1$.

$f(p)\in f(S_1)\subseteq \overline{f(S_1)}$.

Step 3:

Let us suppose that $p\in \overline{S_1}\setminus S_1$.

That means that $p$ is an accumulation point of $S_1$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

$f(p)\in f(S_1)$ (which is possible because $f$ is not supposed to be injective) or $f(p)\notin f(S_1)$.

When $f(p)\in f(S_1)$, $f(p)\in f(S_1)\subseteq \overline{f(S_1)}$.

Let us suppose that $f(p)\notin f(S_1)$ hereafter.

Let any open neighborhood of $f(p)$ be $U_{f(p)}\subseteq T_2$.

As $f$ is continuous, $f^{-1}(U_{f(p)})\subseteq T_1$ is open, and contains $p$, so, is an open neighborhood of $p$.

As $p$ is an accumulation point of $S_1$, there is a point, $p^{\prime }\in f^{-1}(U_{f(p)})\cap S_1$.

$f(p^{\prime })\in U_{f(p)}\cap f(S_1)$, which implies that $f(p)$ is an accumulation point of $f(S_1)$.

So, $f(p)\in \overline{f(S_1)}$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

So, $f(p)\in \overline{f(S_1)}$ anyway.

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3: Note

If $f$ is a homeomorphism, $f(\overline{S_1})=\overline{f(S_1)}$, because $f^{-1}$ is a continuous map and $f^{-1}(f(S_1))=S_1$, and so, $f^{-1}(\overline{f(S_1)})\subseteq \overline{f^{-1}(f(S_1))}=\overline{S_1}$, and so, $f(f^{-1}(\overline{f(S_1)}))=\overline{f(S_1)}\subseteq f(\overline{S_1})$.

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References

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